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Select a random file from a directory and send it (Python, MIME)


By : José Vildósola
Date : October 14 2020, 08:10 PM
I think the issue was by ths following , You select a random file and then throw it away (well, you print it, then throw it away):
code :
files = os.listdir(path)
index = random.randrange(0, len(files))
print(files[index])
filename = files
attachment  = open(filename, 'rb')
attachment  = open(random.choice(files), 'rb')
attachment  = open(os.path.join(path, random.choice(files)), 'rb')


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select random file from directory


By : sarthak
Date : March 29 2020, 07:55 AM
Any of those help select random file from directory
code :
private string getrandomfile2(string path)
    {
        string file = null;
        if (!string.IsNullOrEmpty(path))
        {
            var extensions = new string[] { ".png", ".jpg", ".gif" };
            try
            {
                var di = new DirectoryInfo(path);
                var rgFiles = di.GetFiles("*.*").Where( f => extensions.Contains( f.Extension.ToLower()));
                Random R = new Random();
                file = rgFiles.ElementAt(R.Next(0,rgFiles.Count())).FullName;
            }
            // probably should only catch specific exceptions
            // throwable by the above methods.
            catch {}
        }
        return file;
    }

select random file from directory


By : CDOG
Date : March 29 2020, 07:55 AM
like below fixes the issue Changed to use a single instance of the pseudo-random number generator.
code :
// Use a class variable so that the RNG is only created once.
private Random generator;
private Random Generator
{
    get
    {
        if (this.generator == null)
        {
           this.generator = new Random();
        }
        return this.generator;
    }
}
private string getrandomfile(string path)
{
    string file = null;
    if (!string.IsNullOrEmpty(path))
    {
        var extensions = new string[] { ".png", ".jpg", ".gif" };
        try
        {
            var di = new DirectoryInfo(path);
            var rgFiles = di.GetFiles("*.*")
                            .Where( f => extensions.Contains( f.Extension
                                                               .ToLower() );
            int fileCount = rgFiles.Count();
            if (fileCount > 0)
            {
                int x = this.Generator.Next( 0, fileCount );
                file = rgFiles.ElementAt(x).FullName;
            }
        }
        // probably should only catch specific exceptions
        // throwable by the above methods.
        catch {}
    }
    return file;
}

Select random file from directory


By : Jack
Date : March 29 2020, 07:55 AM
wish of those help You can use glob to get all files in a directory, and then take a random element from that array. A function like this would do it for you:
code :
function random_pic($dir = 'uploads')
{
    $files = glob($dir . '/*.*');
    $file = array_rand($files);
    return $files[$file];
}

PHP load directory select HTML file and send it via Mail


By : bhadram rao
Date : March 29 2020, 07:55 AM
I hope this helps . I set myself to a little PHP project. The aim of my project is to create a webapp which loads HTML emails from directory and send them out. , Try this code
code :
  <?php


function LoadTemplate()
    {
    foreach(glob(dirname(__FILE__) . '/templates/*') as $filename)
        {
        echo "<option value='" . $filename . "'>" . basename($filename) . "</option>";
        }
    }
if(isset($_REQUEST['submit'])){
    $to = $_POST['recipient'];
    $subject = $_POST['subject'];
    $message = file_get_contents($_REQUEST['s1'], "r") or exit("Unable to open file");
    $headers = "Content-type: text/html\r\n";
    $mail_send = mail($to, $subject, $message, $headers);
    if($mail_send){
     echo 'Mail Send ';
     }else{
     echo 'Try Later';
     }
}
?>


<html>
<head></head>
<body>
    <h2>Mail</h2>
    <form name="form1" method="post" action="mail.php">Send To:
        <input type="text" id="recipient" name="recipient">
        <br>Subject:
        <input type="text" id="subject" name="subject">
        <br>
        <select name="s1">
            <option value="" selected="selected">-----</option>
            <?php require( 'mail.php'); echo LoadTemplate(); ?>
            <br>
            <input type="submit" value="Send" name='submit'>
        </select>
    </form>
</body>
</html>

How can I efficiently select 100 random JPG files from a directory (including subdirs) in Python?


By : Кристина Малькова
Date : March 29 2020, 07:55 AM
wish of those help See Generator Tricks for System Programmers for a bunch of neat stuff. But specifically, see the gen-find example. This is as efficient as you are going to get, without making a bunch of assumptions about your file structure layout.
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